Question

初始数据：

select t.AGENHI, t.TACTHI, t.DTEFHI, t.DTFIHI            
      from mytable t
      where agenhi = '81000040' ;

AGENHI      TACTHI   DTEFHI     DTFIHI            
81000040    1        24/02/92   08/03/92
81000040    1        09/03/92   28/02/93
81000040    1        01/03/93   31/05/97
81000040    0,8      01/06/97   31/12/97
81000040    1        01/01/98   31/12/98
81000040    1        01/01/99

使用此查询：

SELECT AGENHI, 
DECODE(TACTHI, 0.05, '005', 0.07, '007', 0.1, '010', 0.137, '013', 0.15, 
'015', 0.2, '020', 0.21, '021', 0.23, '023', 0.25, '025', 0.3, '030', 
0.34, '034', 0.4, '040', 0.45, '045', 0.5, '050', 0.6, '060', 0.63, 
'063',0.7, '070', 0.75, '075', 0.8,'080', 0.84, '084',0.9, '090', 1, '100', 
TACTHI) as QUOTITE,
dtefhi as START_DATE,
DECODE(LEAD (DTFIHI, 1) OVER (ORDER BY DTFIHI NULLS LAST) ,null, 
to_date('31122099','ddmmyyyy'), LEAD (DTFIHI, 1) OVER (ORDER BY DTFIHI NULLS 
LAST)) AS END_DATE
FROM MYTABLE 
WHERE AGENHI = '81000040' AND DTFIHI IS NOT NULL;

我明白了：

AGENHI      QUOTITE     START_DATE    END_DATE
81000040    100         08/03/92      28/02/93
81000040    100         28/02/93      31/05/97
81000040    100         31/05/97      31/12/97
81000040    080         31/12/97      31/12/98
81000040    100         31/12/98      31/12/99

但是当下一行相同时，我需要将“ QUOTITE”分组，并显示第一行的开始日期和最后一行的日期。

预期结果：

AGENHI      QUOTITE     START_DATE    END_DATE
81000040    100         24/02/92      31/12/97
81000040    080         01/06/97      31/12/98
81000040    100         01/01/98      31/12/99

gordon提供的解决方案：

选择agenhi，tacthi，min（dtfihi）作为开始日期，领先（max（dtfihi））over（由agenhi划分，seqnum-seqnum_2由max（dtfihi）排序）作为end_date 从（选择t。*， row_number（）超过（按dtfihi的agenhi顺序划分）为seqnum， row_number（）结束（由agenhi进行分区，由dtfihi进行tacthi排序）为seqnum_2 来自HIA@CHRONOS_TO_S2.WORLD t 其中agenhi ='81000040'并且dtfihi不为空）吨通过agenhi分组（（seqnum-seqnum_2），tacthi；

结果：

81000040    1   08/03/92   null 
81000040    1   31/12/98   null   
81000040    0,8 31/12/97   null

我如何获得结束日期？我将寻找是否找到解决方案，谢谢！

Answer 1

您有一个“孤岛”问题。我将从原始数据开始，像这样：

select agenhi, tacthi, min(dtfihi) as start_date,
       lead(min(dtfihi)) over (partition by agenhi order by min(dtfihi)) as end_date
from (select t.*,
             row_number() over (partition by agenhi, order by dtfihi) as seqnum,
             row_number() over (partition by agenhi, tacthi order by dtfihi) as seqnum_2
      from mytable t
      where agenhi = '81000040' an dtfihi IS NOT NULL
     ) t
group by agenhi, (seqnum - seqnum_2), tacthi;

您可以添加decode()逻辑（我将其写为case表达式），但这并没有真正改变解决方案。

Answer 2

@Boneist和@GordonLinoff再次感谢您的帮助。

select 
agenhi, 
DECODE(TACTHI, 0.05, '005', 0.07, '007', 0.1, '010', 0.137, '013', 0.15, '015', 0.2, '020', 0.21, '021', 0.23, '023', 0.25, '025', 0.3, '030', 
0.34, '034', 0.4, '040', 0.45, '045', 0.5, '050', 0.6, '060', 0.63, '063',0.7, '070', 0.75, '075', 0.8,'080', 0.84, '084',0.9, '090', 1, '100', TACTHI) as QUOTITECPAGE,  
min(dtefhi) as start_date,
nvl(lead(min(dtfihi)) over (partition by agenhi order by min(dtfihi)), 
to_date('31122099','ddmmyyyy')) as end_date
from (select t.*,
         row_number() over (partition by agenhi order by dtfihi) as seqnum,
         row_number() over (partition by agenhi, tacthi order by dtefhi) as seqnum_2
  from HIA@CHRONOS_TO_S2.WORLD t
  where agenhi = '81000040' order by DTEFHI
 ) t
 group by agenhi, (seqnum - seqnum_2), tacthi;

Oracle-从行组获取最小和最大日期

2 个答案: